Question

Use the following data to determine the molar heat of vaporization of chlorine: T (degrees C):...

Use the following data to determine the molar heat of vaporization of chlorine:

T (degrees C): -84.5 -71.2. -47.3
P (mmHg): 40.0 100.0 400.0
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Answer #1

According to Clausius Clapeyron equation :

ln[P] = -(ΔHvap /R)(1/T)+C where P is vapor pressure at temperature T , ΔHvap is heat of vaporization , R is gas constant(=8.314J/mol.K) and C is a constant.

When we draw a graph of ln[P] vs 1/T it will be a straight line and its slope will be equal to -(ΔHvap /R), from which we can get heat of vaporization that is ΔHvap.

The plot obtained is as follows:

Slope of the line obtained is = -2636.3 =  -(ΔHvap /R). So it gives ΔHvap = 2636.3*8.314 = 21918.2 J/mol.

= 21.9 KJ/mol.

Thus molar heat of vaporization of chlorine is 21.9 KJ/mol.

Comment in case of any doubt.

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